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Evaluation of Definite Integrals by Substitution - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

19 Questions around this concept.

Solve by difficulty

The solution for $x$ of the equation $\int_{\sqrt{2}}^{x}\; \frac{dt}{t\sqrt{t^{2}-1}}=\frac{\pi }{2}$ is

A

$\frac{\sqrt{3}}{2}\;$

B

$-\sqrt{2}\;$

C

$\; 2\;$

D

$\; \pi$

Concepts Covered - 1

Evaluation of Definite Integrals by Substitution

We have already learnt to find Indefinite Integration by using substitution method. But in the case of definite integration, we also need to change the limits of integration 'a' and 'b'. If we substitute x = g(t), then g(t) must be continuous in the interval [a, b].

Let's see some solved examples to see how such questions are solved.

Example 1

$\text{Compute the integral } \;\int_{0}^{\pi / 2} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}$

$\\\text { Let } \quad I=\int_{x=0}^{x=\pi / 2} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}\\\text{Divide numerator and denominator by }\cos^2x\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}=\int_{x=0}^{x=\pi / 2} \frac{\sec ^{2} x d x}{a^{2}+b^{2} \tan ^{2} x}\\\text { Put } \quad \text { tan } x=t \Rightarrow \sec ^{2} x d x=d t\\\therefore\;\;\;\; \quad I=\int_{t=0}^{t=\infty} \frac{d t}{a^{2}+b^{2} t^{2}}\\\begin{array}{l}{\text { We find the new limits of integration } t=\tan x \Rightarrow t=0} \\ {\text { when } x=0 \text { and } t=\infty \text { when } x=\pi / 2}\end{array}\\\mathrm{\Rightarrow\;\;\;\;} \quad I=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{d t}{\left(\frac{a}{b}\right)^{2}+t^{2}}=\frac{1}{b^{2}} \cdot \frac{1}{a / b}\left[\tan ^{-1} \frac{b t}{a}\right]_{0}^{\infty}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{a b}\left[\frac{\pi}{2}-0\right]=\frac{\pi}{2 a b}$

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Evaluation of Definite Integrals by Substitution

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